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- Thread starter decamij
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[tex]E_k = \frac{1}{2}m\frac{(2 \pi r)^2}{T^2}[/tex]

With T the period of the orbital motion. Now if you equate the gravitational force to the centripetal force [itex]mv^2/r[/itex] and use the same expression as above for the velocity you get:

[tex]T^2=\frac{(2 \pi r)^2 r}{GM}[/tex]

Wich is Keplers law for circular orbits. Filling this expression in the kinetic energy expression:

[tex]E_k = \frac{GMm}{2r}=-\frac{E_g}{2}[/tex]

This shouldn't be too surprising as total energy is kinetic + potential energy and you already noticed this equalled half the potential energy.

[tex]E_{pot}=E_g=-G\frac{Mm}{r}[/tex]

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